# 2014 Super training #3 H tmutarakan exams-anti-DDoS Principle

Source: Internet
Author: User

Original question: Ural 1091 http://acm.timus.ru/problem.aspx? Space = 1 & num = 1091

Ask for k different numbers so that they have an appointment greater than 1, and all numbers cannot be greater than a specified number S.

Solution: You can consider the prime number in each second. This prime number and all its multiples constitute a set, then you can go to any k elements in these sets, C (n, k) that is, the number of methods in this case, such as K = 3, S = 10,

Three sets can be formed: {2, 4, 6, 8, 10}, {3, 6, 9}, {5, 10}, the first set C (5, 3 ), the second set C (3, 3), and the third set is 0. At the same time, we can see that 6 appears in both sets, so we need to subtract a 6, this is the principle of rejection.

Enumerate one prime number and two prime numbers, because if the three prime numbers are at least 2x3x5 = 30, in this case, there is only one (S <= 50) multiple of 30 in s, that is, 30 itself. It is impossible to select K> = 2. Therefore, three or more items are excluded.

Code:

`#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define ll long longusing namespace std;#define N 10007ll C;int prime = {2,3,5,7,11,13,17,19,23};void calc_C(){    memset(C,0,sizeof(C));    C = C = 1;    for(int i=2;i<=51;i++)    {        C[i] = 1;        for(int j=1;j<=51;j++)            C[i][j] = C[i-1][j] + C[i-1][j-1];    }}int main(){    int k,S,i,j,res,num,flag;    calc_C();    while(scanf("%d%d",&k,&S)!=EOF)    {        ll res = 0;        flag = 9;        for(i=0;i<9;i++)        {            num = S/prime[i];            if(num < k)            {                flag = i;                break;            }            else                res += C[num][k];        }        for(i=0;i<flag;i++)        {            for(j=i+1;j<flag;j++)            {                num = S/(prime[i]*prime[j]);                if(num < k)                    break;                else                    res -= C[num][k];            }        }        printf("%lld\n",min((ll)10000,res));    }    return 0;}`
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